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Chapter 9: Problem 73

A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length \(\ell\) and mass \(m_{1}\) extending radially from the surface of a sphere of radius \(R\) and mass \(m_{2} .\) Find the location of the club's center of mass, measured from the center of the club head.

Short Answer

Expert verified
The position of the center of mass of the golf club, measured from the center of the club head is \( \frac{m_{1} \ell}{2(m_{1}+m_{2})} \)

Step by step solution

01

Define the position vectors

The center of mass of the rod extends from \( \ell /2 \) (due to the uniform distribution of the mass) from the center of the sphere. Similarly, the center of mass of the sphere is at its center, that is 0 from the center of the sphere. Therefore, define the position vectors as: \( r_{1} = \ell /2 \) for the rod and \( r_{2} = 0 \) for the spherical clubhead.
02

Apply the formula for center of mass

The formula for the center of mass \( R \) for a system of particles is given by \( R = \frac{1}{M} \sum_{i} m_{i} r_{i} \) where the \( m_{i} \) are the masses of the individual components, the \( r_{i} \) are the position vectors of the individual components and M is the total mass. In this case, \( m_{1} = m_{1} \), \( m_{2} = m_{2} \), \( r_{1} = \ell /2 \), \( r_{2} = 0 \), and \( M = m_{1} + m_{2} \). Substituting these values into the formula yields: \( R = \frac{1}{m_{1} + m_{2}} \) \( m_{1} \cdot \frac{\ell}{2} + m_{2} \cdot 0 \)
03

Evaluate the expression

Evaluating the expression from step 2 gives the position of the center of mass as \( R = \frac{m_{1} \ell}{2(m_{1}+m_{2})} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Problems
Physics problems involving the center of mass usually demand a clear understanding of how mass distribution affects motion and balance. These problems help in predicting how an object will move when forces are applied. In the context of a golf club modeled as a uniform rod and a sphere, we need to calculate its center of mass to understand how it swings.

To tackle such problems, it's vital to break them down into simple components and analyze the mass and position vector of each part. By doing this, you can simplify the real-life scenario into calculable pieces. This approach is common in physics as it allows us to deal with complex systems smoothly by focusing on fundamental principles.
System of Particles
A system of particles refers to multiple interacting particles, where each particle's properties influence the entire system's behavior. In this golf club problem, the system consists of two particles: the uniform rod (shaft) and the sphere (club head).

To find the center of mass of this system, we must consider both particles' masses and their respective positions. By defining position vectors for each particle, we can use the center of mass formula, which integrates all parts into one weighted average location.
  • The rod and the sphere are the system's components in this scenario.
  • The position of each part is crucial for determining the overall center of mass.
  • Each mass contributes differently depending on its mass and location.
Understanding systems of particles is essential because many real-world problems in physics involve analyzing how groups of objects interact to produce a specific outcome.
Position Vectors
Position vectors are a mathematical tool used to describe the location of an object relative to a reference point. In physics problems like this one, position vectors help us determine where each part of our system resides in space.

The center of mass of the rod is at its midpoint due to its uniformity, with a position vector described as \( r_{1} = \ell /2 \). This means the mass of the rod is evenly spread around this point. For the sphere, since its entire mass is symmetrically distributed, the position vector is \( r_{2} = 0 \), directly in the center of the club head.
  • Position vectors allow us to simplify and solve complex physical problems.
  • They provide a way to measure each part's contribution to the entire system.
Utilizing position vectors simplifies finding the center of mass, making them an indispensable tool in physics.
Uniform Rod
A uniform rod is a rod with a constant mass distribution along its length. In our problem, this refers to the golf club shaft. Because the rod is uniform, its center of mass lies exactly in the middle, or \( \ell /2 \) from either end.

When solving problems involving uniform rods, it's crucial to recognize that each small segment of the rod contributes equally to its overall mass. Therefore, its center of mass is simply the midpoint.
  • Uniformity implies an even distribution of mass along the entire length.
  • The rod contributes significantly to the overall center of mass of the system.
  • Recognizing uniform distribution simplifies many calculations in physics.
Incorporating the properties of a uniform rod into our calculations helps in efficiently predicting and understanding the physical behavior of systems like the golf club.

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Most popular questions from this chapter

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed \(v_{i} .\) After the collision, the orange disk moves along a direction that makes an angle \(\theta\) with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.

A \(0.100-\mathrm{kg}\) ball is thrown straight up into the air with an initial speed of \(15.0 \mathrm{m} / \mathrm{s} .\) Find the momentum of the ball (a) at its maximum height and (b) halfway up to its maximum height.

In a slow-pitch softball game, a \(0.200-\mathrm{kg}\) softball crosses the plate at \(15.0 \mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ}\) below the horizontal. The batter hits the ball toward center field, giving it a velocity of \(40.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal. (a) Determine the impulse delivered to the ball. (b) If the force on the ball increases linearly for \(4.00 \mathrm{ms}\), holds constant for \(20.0 \mathrm{ms},\) and then decreases to zero linearly in another \(4.00 \mathrm{ms},\) what is the maximum force on the ball?

A friend claims that, as long as he has his seat belt on, he can hold on to a 12.0 -kg child in a \(60.0 \mathrm{mi} / \mathrm{h}\) head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.050 0 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.

A 45.0 -kg girl is standing on a plank that has a mass of 150 kg. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The girl begins to walk along the plank at a constant speed of \(1.50 \mathrm{m} / \mathrm{s}\) relative to the plank. (a) What is her speed relative to the ice surface? (b) What is the speed of the plank relative to the ice surface?
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